The problemCan be ζ(3) written as απβ, where (α,β∈C), β≠0 and α doesn’t depend of π (like √2, for example)?

DetailsSeveral ζ values are connected with π, like:

ζ(2)=π2/6

ζ(4)=π4/90

ζ(6)=π6/945

…

and so on for all even numbers.

See this mathworld link to more details: Riemann Zeta Function

So the question is, could ζ(3) be written as:

ζ(3)=απβ

α,β∈C

β≠0

α not dependent of πSee α not essencially belongs Q and α,β could be real numbers too.

When I wrote α is not dependent of π it’s a strange and a hard thing to be defined, but maybe α can be written using e or γ or √2 or some other constant.

Edit:Maybe

this still a open question. If∑2k=0(−1)kB2k B2−2k+2(2k)! (2−2k+2)!

in −4∑2k=0(−1)kB2k B2−2k+2(2k)! (2−2k+2)!π3 be of the form δπ3 with δ

not dependentof πand −2∑k≥1k−3e2πk−1

not dependentof π too, this question still hard and open.

Edit 2:I discovered a result, but later I’ve seen that this is something already known, either way, it is an interesting one to have it here.

ζ(3)=−4π2ζ′(−2)

but, if ζ′(−2) is of the form απ2, with α not dependent of π, then this still remains as a hard and an open question.

I have a conjecture that ζ′(−2) will not cancel the π2 term, but since I wasn’t able to prove it and I can’t use it here.

We can express ζ of odd numbers with ζ′ in a easy way, with a “closed” form like this one.

**Answer**

The question is whether or not ζ(3) is *connected* with π. The answer is *yes*. Moreover, ζ(3)=βπα for some complex α,β. Take α=3 and β=0.0387682…. It is not known whether β=0.0387682… is algebraic or transcendental. It is known, however, that ζ(3) is irrational as shown by Apery.

Ramanujan conjectured and Grosswald proved that the following holds. If α,β>0 such that αβ=π2, then for each non-negative integer n,

α−n(ζ(2n+1)2+∑k≥1k−2n−1e2kα−1)=(−β)−n(ζ(2n+1)2+∑k≥1k−2n−1e2kβ−1)−

22nn+1∑k=0(−1)kB2k B2n−2k+2(2k)! (2n−2k+2)!αn−k+1βk.

where Bn is the nth-Bernoulli number.

For odd positive integer n, we take α=β=π,

ζ(2n+1)=−22n(n+1∑k=0(−1)kB2k B2n−2k+2(2k)! (2n−2k+2)!)π2n+1−2∑k≥1k−2n−1e2πk−1.

In particular, for n=1,

ζ(3)=−4(2∑k=0(−1)kB2k B2−2k+2(2k)! (2−2k+2)!)π3−2∑k≥1k−3e2πk−1.

Observe that the coefficient of π3 is *rational*, however, nothing is known about the algebraic nature of the infinite sum. This is a current topic of research. Indeed, it is conjectured that ζ(3)π3 is *transcendental*.

*Update*: Recently, Takaaki Musha claims to have proved that ζ(2n+1)(2π)2n+1 is irrational for positive n≥1. However, some objection has since been raised (read comments below).

**Attribution***Source : Link , Question Author : GarouDan , Answer Author : user02138*